@Don- Yes, I agree the choice of Random number generator algorithm should
very much depend on the underlying application.
Thanks, for bringing up the Diehard tests, those are some interesting
statistical tests. +1
On 26 October 2012 21:23, Don dondod...@gmail.com wrote:
How you chose from the
How you chose from the wide array of pseudo-random generators
available depends a lot on how you intend to use it. If you just need
something that seems random in a video game, a LCG is probably fine.
However, you could fill a large bookshelf with studies which were
invalidated because they used
Mersenne Twister is more than sufficient for simulation or most other
applications other than encryption.
Don
On Oct 25, 6:37 am, Anuj Khandelwal anuj.cool.khandel...@gmail.com
wrote:
hey all,
Any idea to generate random number without using rand() function call ?
Any algorithms for random
I gave this method as you are also expecting random(0,1) to give values
either 0 or 1 only, not between these two.
Thanks and Regards
Devansh Gupta
MNNIT, Allahabad
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Yes, Random(0, 1) gives values 0 or 1 only with equal probabilities. But
your solution won't work.
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if b-a is exactly 2^k-1 , k0
then a + k bits (each bit is set using rand(0 1) ) with equal probability
surender
On Thu, Jul 7, 2011 at 1:20 PM, Nitish Garg nitishgarg1...@gmail.comwrote:
Yes, Random(0, 1) gives values 0 or 1 only with equal probabilities. But
your solution won't work.
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You
Thanks Dave, nice solution. By setting the bits we can get equal
probabilities.
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@Nitish: I'm assuming that Random(0,1) returns 0 and 1 randomly, with
equal probability.
Let n = ceiling(log_2(b-a+1)).
Use the rejection method (http://en.wikipedia.org/wiki/
Rejection_sampling) as follows:
Generate an integer, k, 0 = k 2^n, one bit at a time, using
Random(0,1) n times. k will
@surender
Yep, I think the same.
for(i=0;i(b-a);i++,a+=rand(0,1));
Watsay?
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Does not look good to me, Here's why :
3+rand(0,1)+rand(0,1)+rand(0,1) : When we have rand(0, 1) + rand(0, 1) +
rand(0, 1), a number of possibilities apply:
We will have the following cases : 0, 0, 0 or 0, 0, 1 or 0, 1, 0 or 0, 1, 1
or 1, 0, 0 or 1, 0, 1 or 1, 1, 0 or 1, 1, 1, their respective
call function and save each of the results in x and y(say)
if x = 0 and y = 1 ...prob = (.6)*(.4) return 0
if x = 1 and y = 0 ...prob = (.4)*(.6) return 1
otherwise, repeat..
hence this new function will return 0 and 1 with 50% prob.
On Tue, Mar 22, 2011 at 11:07 PM, Don dondod...@gmail.com
There is no upper bound on the runtime of this function.
On Mar 20, 4:33 pm, Jammy xujiayiy...@gmail.com wrote:
clearly the prob. of getting true|false and false|true are equal to
0.6*0.4. Therefore the following code works,
bool uniform(){
bool f1;
bool f2;
do{
f1 =
bool ff()
{
while(true) {
int a = f() ? 2 : 0;
int b = f() ? 0 : -1;
if (a + b == 1) return true;
if (a + b == 0) return false;
}
}
On Wed, Mar 23, 2011 at 1:54 AM, Don dondod...@gmail.com wrote:
There is no upper bound on the runtime of this function.
i dont know how to write a function which returns true 60% time and
false 40% time
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clearly the prob. of getting true|false and false|true are equal to
0.6*0.4. Therefore the following code works,
bool uniform(){
bool f1;
bool f2;
do{
f1 = non_uniform();
f2 = non_uniform();
}while(!(f1 ^ f2));
return f1;
}
On Mar 17, 10:24 am, saurabh agrawal
@gene: can u give the function for unfair_coin_toss so that i can
understand what u are doing in fair_coin_toss
On Mar 19, 9:54 am, Gene gene.ress...@gmail.com wrote:
On Friday, March 18, 2011 1:47:45 PM UTC-4, Gene wrote:
On Mar 17, 10:24 am, saurabh agrawal saura...@gmail.com wrote:
On Saturday, March 19, 2011 6:07:44 AM UTC-4, cegprakash wrote:
@gene: can u give the function for unfair_coin_toss so that i can
understand what u are doing in fair_coin_toss
On Mar 19, 9:54 am, Gene gene.r...@gmail.com wrote:
On Friday, March 18, 2011 1:47:45 PM UTC-4, Gene wrote:
On Mar 17, 10:24 am, saurabh agrawal saurabh...@gmail.com wrote:
Given a function which returns true 60% time and false 40% time.
Using this function you have to write a function which returns true 50% of
the time.
If we call the function N times, then the probability that we'll have
K true
On Friday, March 18, 2011 1:47:45 PM UTC-4, Gene wrote:
On Mar 17, 10:24 am, saurabh agrawal saura...@gmail.com wrote:
Given a function which returns true 60% time and false 40% time.
Using this function you have to write a function which returns true 50%
of
the time.
If we
bool uniform()
{
static bool state = true;
state = !state;
return state ^ nonuniform();
}
On Mar 17, 9:24 am, saurabh agrawal saurabh...@gmail.com wrote:
Given a function which returns true 60% time and false 40% time.
Using this function you have to write a function which returns true
@Don: Clever and efficient code to meet the problem statement
precisely. It does give true on half of the calls, but it doesn't give
true with probability 50% on any given call, so we wouldn't want to
use it to model a fair coin.
Dave
On Mar 17, 1:03 pm, Don dondod...@gmail.com wrote:
bool
@harit
random number means its output cant be predicted,it doesnt mean that the
probability will be equal for any number of test runs.
e.g consider random number between 3-5 for 3 test runs
if the first output is 4
second is 3
then to ensure same probability third output must be 5
which makes
okay thanks
On Sep 20, 2:59 am, Gene gene.ress...@gmail.com wrote:
On Sep 18, 5:45 pm, eSKay catchyouraak...@gmail.com wrote:
okay now I get it.
but then
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now
How are you picking ni from [a...b] range (length 7) ?
From: avalon avalo...@gmail.com
To: Algorithm Geeks algogeeks@googlegroups.com
Sent: Saturday, 19 September, 2009 2:16:47 PM
Subject: [algogeeks] Re: random number...
Forgive me first if i am wrong since
Isn't it the same of adding two random numbers
When you just add two random numbers, you tend to normalize them. If
you simply multiply one
of them by 5 before adding with the other, you'll end up normalizing
them the same way...
On 18 set, 18:45, eSKay catchyouraak...@gmail.com wrote:
Walter, when you roll two dice, you get two random numbers between 1
and 6 inclusive. Each has an essentially uniform distribution, but the
sum does not. Of every 36 rolls, you will average one 2, two 3s, three
4s, four 5s, five 6s, six 7s, five 8s, four 9s, three 10s, two 11s,
and one 12. The
okay now I get it.
but then
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now uniformly random
between 1 and 25
} while(i 7);
return i;
should be fine. isn't it??
Random number between any two numbers (a,b) can also be obtained as
follows.
let rand() return a random number between 0 and 1 ;
int random(a,b)
{
if(a==b)
return a ;
mid = a + (b-a)/2 ;
if( rand() )
return random(mid,b) ;
else
return random(a,mid) ;
}
I think probability of each
On Sep 18, 5:45 pm, eSKay catchyouraak...@gmail.com wrote:
okay now I get it.
but then
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now uniformly random
between 1 and 25
} while(i 7);
return i;
On Sep 19, 4:46 am, avalon avalo...@gmail.com wrote:
Forgive me first if i am wrong since i didn't read all the posting ...
Here is a way to sol the problem.
n1 = random_1_5() + 0;
n2 = random_1_5() + 5;
..
n7= random_1_5() + (7-1)*5;
now n1 ... n7 is in range [1 ... 35]
Image we divde
one of the solutions given at
http://stackoverflow.com/questions/137783/given-a-function-which-produces-a-random-integer-in-the-range-1-to-5-write-a-fun
is:
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now uniformly random
What is wrong with using i = rand5() + rand5() instead of i = 5 * rand5
() + rand5() is that the former is not uniformly distributed between 1
and 10, as claimed. First, 1 never occurs. 2 occurs 1 out of 25 times,
3 occurs 2 out of 25, etc. (Think of rolling two ordinary dice.)
Dave
On Sep 18,
On Sep 7, 11:56 am, ankur aggarwal ankur.mast@gmail.com wrote:
Given a random number generator that generates numbers in the range 1 to
5, how can u create a random number generator to generate numbers in the
range 1 to 7. (remember that the generated random numbers should follow a
dave_and_da...@juno.com
To: Algorithm Geeks algogeeks@googlegroups.com
Sent: Wednesday, 9 September, 2009 2:42:26 AM
Subject: [algogeeks] Re: random number...
Manish, your first approach doesn't work. You will notice that b1, b2,
and b3 each are 0 2/5 of the time and 1 3/5 of the time, so the return
value
@dave
*
*On Tue, Sep 8, 2009 at 7:27 AM, Dave dave_and_da...@juno.com wrote:
Use the rejection technique, something like this:
do
{
do
U1 = random_1_to_5();
while( U1 == 5 );
// At this point, U1 is a uniform integer in the range 1 to 4.
U2 = random_1_to_5();
* if(
Hardest part of this problem is to prove that the generated number
INDEED follow uniform distribution :D
_dufus
On Sep 8, 6:57 am, Dave dave_and_da...@juno.com wrote:
Use the rejection technique, something like this:
do
{
do
U1 = random_1_to_5();
while( U1 == 5 );
// At
@dufus
tell me 1 thing
do we have to make algo so that the prob is 1/7 or do we have to make so
that prob of generating number between 1 to 7 is same may be( 1/10) etc ???
On Tue, Sep 8, 2009 at 11:42 AM, Dufus rahul.dev.si...@gmail.com wrote:
Hardest part of this problem is to prove that the
Your comment is moot, since if any random number generator returns the
same number forever, you are not going to be able to use it to
generate other random numbers with a uniform density.
On Sep 8, 9:04 am, Karthik Singaram Lakshmanan
karthiksinga...@gmail.com wrote:
The rejection
@karthik
but look at the prob part..
it seems ok 2 me..
it is random generator u cannot guarntee which number is going 2 come..
On Tue, Sep 8, 2009 at 7:34 PM, Karthik Singaram Lakshmanan
karthiksinga...@gmail.com wrote:
The rejection technique may never halt...(with an infinitesimally
@dave
plz quote the name of the person u r taking / pointing..
On Tue, Sep 8, 2009 at 8:11 PM, Dave dave_and_da...@juno.com wrote:
Your comment is moot, since if any random number generator returns the
same number forever, you are not going to be able to use it to
generate other random
On Tue, Sep 8, 2009 at 9:04 PM, ankur aggarwal ankur.mast@gmail.comwrote:
@dave
plz quote the name of the person u r *talking* / pointing..
On Tue, Sep 8, 2009 at 8:11 PM, Dave dave_and_da...@juno.com wrote:
Your comment is moot, since if any random number generator returns the
same
Huh? You'll note his posting is appended to mine, just as your
posting, my posting, and part of his posting are appended to this one.
On Sep 8, 10:34 am, ankur aggarwal ankur.mast@gmail.com wrote:
@dave
plz quote the name of the person u r taking / pointing..
On Tue, Sep 8, 2009 at
Ramaswamy, summing seven uniform random numbers doesn't give a uniform
result. Even the sum of two random numbers is not uniform. Consider
rolling two dice. 2 occurs only 1 out of 36 times, while 7 occurs 6
out of 36 times.
Dave
On Sep 8, 2:34 pm, Ramaswamy R ramaswam...@gmail.com wrote:
Manish, your first approach doesn't work. You will notice that b1, b2,
and b3 each are 0 2/5 of the time and 1 3/5 of the time, so the return
value is not uniformly distributed.
For approach 2, what do you have to do if you want an exactly uniform
distribution as a result? Or what would the code
Generate the random number 7 times. Sum them all and divide by 5.
Theoritically it should be evenly distributed over 1-7. But owing to random
number generators characteristics the sum of rand(5) called 7 times may not
be perfectly evenly distributed over 1-7.
A nice discussion on some neat options
Not that I am trying to prove a point...but just trying to clarify
what I said...
A uniform random number generator,
say rand_5() produces a value 2 with probability 1/5
it can produce a sequence of two 2s with probability (1/5)^2
it can produce a sequence of three 2s with probability (1/5)^3
I KNow a sol for given a rand_5() function which generates *0* to 5
and we have to find rand_7() for *0* to 7
could not think about it...
On Mon, Sep 7, 2009 at 9:26 PM, ankur aggarwal ankur.mast@gmail.comwrote:
Given a random number generator that generates numbers in the range 1 to
@ankur: u r right.
On Mon, Sep 7, 2009 at 9:36 PM, ankur aggarwalankur.mast@gmail.com wrote:
I KNow a sol for given a rand_5() function which generates 0 to 5
and we have to find rand_7() for 0 to 7
could not think about it...
On Mon, Sep 7, 2009 at 9:26 PM, ankur aggarwal
Use the rejection technique, something like this:
do
{
do
U1 = random_1_to_5();
while( U1 == 5 );
// At this point, U1 is a uniform integer in the range 1 to 4.
U2 = random_1_to_5();
if( U1 2 )
U2 += 5;
}
while( U2 7 );
// At this point, U2 is a uniform random
Monu Rathour wrote:
i worked in *visual studio 8*,and there is a function *rand()* to generate
random numbers.
But now i have to work on *visual studio 6*, is there any such function?
rand() is the function provided by the standard C library (stdlib), so
as long as you're using an ANSI C
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